I have the equation $(1)$ defined on $x \in [0,L]$ and $y \in [0,d]$
$$ <\nabla^2-\beta><\mathcal{D}>T_2(x,y)+\alpha\beta T_2(x,y)=-\alpha m\gamma \frac{\cosh(m(d-y))}{\sinh(md)} \tag 1 $$ where $$<\mathcal{D}>=\frac{\partial}{\partial x} + \alpha$$ and $$<\nabla^2-\beta>=\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}-\beta$$ governed by the boundary conditions $$ \frac{\partial T_2}{\partial x} \vert_{x=0} = \frac{\partial T_2}{\partial x} \vert_{x=L} = \frac{\partial T_2}{\partial y} \vert_{y=d} = \frac{\partial T_2}{\partial y} \vert_{y=0}=0 \tag 2 $$ Also the following is known $$ <\nabla^2-\beta>T_2 (0,y)=0 \tag 3 $$
How should I approach to solve $(1)$ ?
Trying an ansatz like $ T_2(x,y)=\sum_{k=0}^{\infty} f_k(x)\cos(\frac{k\pi y}{d}) \tag 9 $ over-complicates thing. Can there be any other approach ?